Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following equals? $$ \lim_{x \to 1}\frac{\displaystyle\int_1^x \sin(t) \, dt}{x^2-1} $$ I think this can be converted to $$ \lim_{x \to 1}\frac{\sin(x)}{2x} = \frac{\sin(1)}{2} $$ by using the fundamental theorem of calculus.
But the correct answer is $1/2$.
So where I made a mistake?

share|improve this question
1  
Do you mean this? $$\lim_{x \to 1} \frac{\int_1^x \sin t dt}{x^2 - 1}$$ –  T. Bongers Apr 12 at 22:15
    
Yes..I don't know how to type it that way sorry –  shidangai Apr 12 at 22:15
3  
You can definitely use l'Hôpital's theorem and the fundamental theorem of calculus; your computation seems good. –  egreg Apr 12 at 22:17

1 Answer 1

up vote 4 down vote accepted

What you've computed (apparently with L'Hospital's rule) is correct. For an alternative way, use the definition of the derivative: we start by factoring the denominator to compute

$$\lim_{x \to 1} \frac{1}{x + 1} \cdot \frac{\int_1^x \sin t}{x - 1}$$

Computing each limit separately, the limit is equal to

$$\frac 1 {1 + 1} \cdot \frac{d}{dx} \left(\int_1^x \sin t dt\right)\Big|_{x = 1} = \frac 1 2 \sin 1$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.